HeatTransfer_3

7/29/2019 HeatTransfer_3

1/17

3/28/2013ES 312 Energy Transfer Fundamentals 116

Two-Dimensional Heat Transfer

Thus far, we have only considered cases where heatfluxes (or temp. variations) are in one main direction.

Lets go beyond this and consider cases of 2-D heattransfer in both simple and complex geometries.

We wont do 3-D, but it uses the same methods as 2-D - just harder.

The governing equation for multidimensional, steadyheat conduction is:

Before we begin to consider how to solve this, lets

first just consider the properties of the Laplacianoperator.

k

q

T

2


2/17

3/28/2013AE 301 Aerodynamics I 117

The Laplacian Operator

In 2-D Cartesian Coordinates, the Laplacian operatoris nothing more than:

This operator is a great smoothing function

The solution of this equation is smooth and continuous,except possibly at discontinuous boundaries.

The maxima and minima of the solution occur on theboundaries (if = 0)

This equation represents many common physical problemssuch as electromagenetic flux, gravitational potential,inviscid and incompressible flow, and the deflection of elasticmembranes.

As a result, this equation has been studied for many years.

k

q

y

T

x

TT

2

2

2

22

q


3/17


The Laplacian Operator (cont)

The book considers 3 ways of solving this equation. Analytic mathematical solutions:

Fairly easy to obtain for simple geometries and simple BCs.

A horror to find on complex geometries and/or BCs.

Gives complete, continuous solutions of T(x,y)

Graphical solutions:

Approximate method which exploits smoothness feature.

Relatively painless - but more of an art than a science.

Numerical solutions:

Hard to program, but once the code exists, pretty easy touse for many different cases.

Only provides solutions of T and discrete points.


4/17


Analytic 2-D Solutions

Consider the simply geometrical case as shown - aretangular region the same temperature on 3 sides,but different on the fourth.

To solve this problem, we use the technique of

Separation of Variables. First, lets simplify the BCsby defining

With this change, our governing eqn.

and BCs for no heat generation are:

02

2

2

2

yx

12

1),(),(TT

TyxTyx

0)0,( x

1),( Wx

0),0( y

0),( yLL

W

T1

T2

T1 T1

x

y


5/17


Analytic 2-D Solutions (cont)

Now, assume that the solution will have the form:

So that, after substituting into the governing eqn:

Or

Where must be a constant since the first term isonly a function of x and the second is only a functionof y!

)()(),( yYxXyx

0

)(

)(

)(

)( 2

2

2

2

dy

yYd

xXdx

xXd

yY

2

2

2

2

2 11

dy

Yd

Ydx

Xd

X


6/17



Thus, the single 2nd order PDE becomes two 1storder ODEs:

Which have the general solutions for X(x) and Y(y):

Or, the general solution with BCs is:

022

2

Xdx

Xd 0

2

2

2

Ydy

Yd

)sin()cos()( 21 xCxCxX yy eCeCyY 43)(

yy eCeCxCxCyx 4321

)sin()cos(),(

0)0,( x

1),( Wx

0),0( y

0),( yL


7/17



To satisfy the first BC, , it must be true thatC1 = 0.

To satisfy the second, , we must have C4 =-C3

To satisfy the second, , the product Lmust be a multiple of. Or

Thus, we dont have a single solution, but a whole

series of them which can be superimposed:

1

32 sin),(n

Lyn

Lyn

nn eeL

xnCCyx

0),0( y

0),( yL

4.....3,2,1, nL

n

0)0,( x


8/17



To simplify this a little, lets introduce the sinhfunction and group all the constants into a singleterm:

To get:

To get this last constant (or series of constants!) weapply the final BC:

L

yn

L

xnCyx

n

n sinhsin),(1

Lyn

CeeCC nLyn

Lyn

nn sinh32

1),( Wx

L

Wn

L

xnCWx

n

n sinhsin1),(1


9/17



This last equation appears impossible to solve.However, there is a little mathematical trick.

Lets take this equation, multiply both sides bysin(mx/L) and integrate from 0 to L.

The trick is that the integral on the RHS is zeroUNLESS m=n, when it becomes L/2. This result is

related to the concept of orthogonal functions. Thus:

L

n

n

L

dxLxm

Lxn

LWnCdx

Lxm

010

sinsinsinhsin

L

WnLCdx

L

xn nL

sinh

2

sin

0


10/17



Finally, the integral on the LHS will be zero for evenvalues of n. For odd values it has the solution 2L/n.

Or, playing algebra games:

And our final solution for this problem is:

7....5,3,1,sinh2

2

n

L

WnLC

n

L n

4....3,2,1,sinh21)1( 1

n

L

Wn

nC nn

LWn

Lyn

L

xn

nyx

n

n

/sinh

/sinhsin

1)1(2),(

1

1


11/17



All this effort. And this was a simple problem! Also, while this solution has the value of being

continuous in x and y, it also involves an infinite sum.

In practice, we might only include the first 10 or so

terms, but where we truncate thesum can effect the solution.

By the way, if we evaluatethis sum, the solution for

(x,y) looks something like:

=1

x

y

=0

=0

=0

0.75

0.5

0.25

0.1


12/17


Graphical Method for 2-D Conduction

The second method of 2-D conduction analysis iscalled the Graphical Method.

In this method, isotherms (lines of constant T) andadiabats (line of constant heat flux) are sketched

based on simple geometric rules and a desire toachieve smoothness.

As a result, this method is approximate - but it canbe fairly accurate if the artist has experience.

The method is also relatively quick - an advantage ifonly preliminary results are needed and theboundaries have known temperatures.


13/17


Graphical Method for 2-D Conduction

Begin any problem by identifying lines of symmetry -which should also be adiabats!

Begin any problem byidentifying lines of symmetry

- which should also beadiabats!

Next sketch in smoothisothermal lines, making sure

they are perpendicular to theadiabats.

T2

T1

Lines ofsymmetry

Isotherms


14/17


Graphical Method (cont)

Draw adiabats in as lines perpendicular at everycrossing with the isotherms.

22

bdacy

cdabx

Finally begin iterating on the linestrying to obtain equal arc distances

on opposing sides using:

When finished, you have a graphical(qualitative) idea of the heat fluxpatterns.

adiabats

a b

c

d


15/17


Graphical Analysis (cont)

To get quantitative results from the graphicaldiagram, use the following procedure.

If properly drawn, there will be M lanes (spacebetween adiabats) of equal heat heating rate.

Similarly, there will be N temperature steps (spacebetween isotherms) of equal temperatureincrements.

If the plot was drawn correctly, with balanced side

lengths, then the total heat flux is just:

where is the width in the 3rd dimension.

21 TkN

M

l

q

l


16/17


The Conduction Shape Factor

A simply way to represent analytical or graphicalresults for 2-D (or 3-D) heat conduction is in the formof a shape factor, S:

This representation depends upon having two welldefined surface temperatures, T1 and T2.

Also not the the thermal resistancefor this case is just:

Some common shape factors are giving in Table 3-1of the text.

The process is simple: find the desired geometry,calculate S, find q or R from the above eqns.

21 TSkq

SkR condt

1,


17/17


Numerical Solutions

The governing equation for multidimensional heatconduction may also be solved using a number ofdifferent numerical solutions.

The book discusses the classical Finite Differencesolution method

However, the FEHT program provided uses the FiniteElement method of analysis.

We will not worry about how to set up the solutions,just how to apply the program to different cases andconditions.

k

qT

2

Documents

HeatTransfer_3